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Solving Inequalities, Compound Inequalities and Absolute Value Inequalities
Sec 1.5 &1.6
pg. 33 - 43
Hallford © 2007 Glencoe © 2003
Objectives
The learner will be able to (TLWBAT): solve inequalities solve real-world problems with
inequalities solve compound inequalities solve absolute value inequalities
Hallford © 2007 Glencoe © 2003
Properties
Trichotomy Property For any two real numbers a and b,
a < b a = b a > b
Hallford © 2007 Glencoe © 2003
Properties of Inequalities
For any real number a, b, and c Addition Prop.
If a>b, then a + c > b + c If a<b, then a + c < b + c
Subtraction Prop. If a>b, then a – c > b – c If a<b, then a – c < b – c
See pg. 33
Hallford © 2007 Glencoe © 2003
Work this problem
x – 7 > 2x + 2 -x -x -7 > x + 2 -2 -2 -9 > x Now we need to graph our answer on
a number line.
Hallford © 2007 Glencoe © 2003
Graphing on a Number Line
Let’s look at our previous answer, -9 > x
-9 0
What does this answer mean?
Are my “x’s” here?Or here?
-9 is GREATER than x, so my “x’s” that make senseare -10, -10.1, -11, etc. Anything smaller than -9
For < or > you use an opening dot or point -
For < or > you use a closed dot or point -
Hallford © 2007 Glencoe © 2003
Properties of Inequalities
Multiplication Prop. For any real numbers
a, b, and c If c is positive If a>b, then ac>bc If a<b, then ac<bc
If c is negative If a>b, then ac<bc If a<b, then ac>bc
Division Prop. For any real numbers
a, b, and c If c is positive If a>b, then a/c>b/c
If a<b, then a/c<b/c
If c is negative If a>b, then a/c<b/c
If a<b, then a/c>b/c
Hallford © 2007 Glencoe © 2003
Work this problem
3x – 7 < 7x + 13
-4x – 7 < 13
7 7
-4x < 20
-4 -4
x > -5
Now let’s graph
-7x -7x
0-5
You can also use set builder notation toexpress your answer.
{x | x > -5}
This is read as the set of all x such that x is greater than or equal to -5
Hallford © 2007 Glencoe © 2003
Work this problem
3(a +4) – 2(3a +4) ≤ 4a - 1
3a + 12 – 6a – 8 ≤ 4a – 1
-3a + 4 ≤ 4a – 1
4 ≤ 7a - 1
5 ≤ 7a5/7 ≤ a
Now graph andexpress in setbuilder notation.
0
{a | a ≥ 5/7}
You can also use interval notation. Interval notation uses ( & ) for < or > and [ & ] for ≤ & ≥. We also use -∞ (negative infinity) & +∞ (positive infinity)
Interval notation for this problem would be [5/7, +∞)
Hallford © 2007 Glencoe © 2003
Compound Inequalities
A compound inequality consists of two inequalities joined by the word “and” or the word “or”.
You must solve both inequalities and then graph. The final graph of the “and” inequality is the
intersection of both individual solution graphs. The final graph of the “or” inequality in the union
of both individual solution graphs
Hallford © 2007 Glencoe © 2003
Let’s solve an “and” compound inequality
7 < 2x - 1 < 15
Method 1 – Divide into two problems
7 < 2x – 18 < 2x4 < x
Method 2 – Work together
2x – 1 < 152x < 16x < 8
7 < 2x – 1 < 158 < 2x < 164 < x < 8
Whateveryou do toone sidedo to the other!
Now Graph
4 < x
x < 8
0
0
0
4
4
4
8
8
8
{ x | 4 ≤ x < 8}
“Means 7 < 2x -1 and 2x – 1< 15”
4 < x < 8
Hallford © 2007 Glencoe © 2003
Let’s work about the same problem as an “or” inequality
7 < 2x -1 or 2x – 1 < 15
We know the answer is 4 < x and x < 8, but thistime the answer graph is different. It is the UNIONof the two graphs.
4 < x
x < 8
0
0
0
4
4
8
8
The solution set is all real numbers .
4 < x or x < 8
Hallford © 2007 Glencoe © 2003
Work this problem
2x + 7 < -1 or 3x + 7 > 10
2x + 7 < -1 3x + 7 > 10
2x < -8x < -4
3x > 3x > 1
0-4
0
or
1
0-4 1
Hallford © 2007 Glencoe © 2003
Absolute Value Inequalities
|a| < b, where b > 0 work as an “and” problem -b < a < b
|a| > b, where b > 0 work as an “or” problem a > b or a < -b
Hallford © 2007 Glencoe © 2003
Work these problems
|x – 1| < 3 -3 < x -1 < 3 -2 < x < 4
|x -1 | > 3 x -1 > 3 or x -1 < -3 x > 4 or x < -2
0 0
Hallford © 2007 Glencoe © 2003
Absolute Value Inequalities
|a| < b, where b < 0 if b is less than zero, it is negative, so
there is no solution |a| > b, where b < 0
if b is less than zero, it is negative, so every real number is a solution or all reals.
Hallford © 2007 Glencoe © 2003
Work these problems
|2x – 7| < -5
There is no solutionfor the above sincethe absolute value cannot be less than zero.
|3x – 1| + 9 > 2
|3x – 1| > -7
Any value of “x” willmake this statementtrue, since the absolute value is alwaysgreater than a negativenumber
All Real numbers